College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 1 - Section 1.8 - Absolute Value Equations and Inequalities - 1.8 Exercises: 37

Answer

$\left[ -\dfrac{2}{3},4\right]$

Work Step by Step

$\bf{\text{Solution Outline:}}$ To solve the given inequality, $ \left|5-3x \right|\le7 ,$ remove the absolute value sign using the properties of absolute value inequality. Then use the properties of inequality to isolate the variable. $\bf{\text{Solution Details:}}$ For any $a\gt0,$ $|x|\le a$ implies $-a\le x\le a.$ (Note that the symbol $\le$ may be replaced with $\lt.$) Hence, the inequality above is equivalent to \begin{array}{l}\require{cancel} -7\le5-3x \le7 .\end{array} Using the properties of inequality, the inequality above is equivalent to \begin{array}{l}\require{cancel} -7-5\le5-5-3x \le7-5 \\\\ -12\le-3x \le2 .\end{array} Dividing both sides by $ -3 $ and reversing the inequality symbols, the inequality above is equivalent to \begin{array}{l}\require{cancel} \dfrac{-12}{-3}\ge\dfrac{-3x}{-3} \ge\dfrac{2}{-3} \\\\ 4\ge x \ge-\dfrac{2}{3} \\\\ -\dfrac{2}{3}\le x \le4 .\end{array} Hence, the solution is the interval $ \left[ -\dfrac{2}{3},4\right] .$
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